20061ee131A_1_131A1sol

20061ee131A_1_131A1sol - is correct. P (correct order) = 1...

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EE 131A Homework #1 Winter 2006 Solution K. Yao 1. The number of distinct ordered 5-tuples with replacements = 10 3 × 5 × 2 = 10 4 . 2. The number of digits greater than 6 is 3 and the number of digits less or equal to 6 is 7. Thus, the total number is 3 × 7 + 7 × 3 = 42. 3. On the first day, there is a choice of 5 pairs, and subsequently, there is a choice of only 4 pairs. Total number is = 5 × 4 × 4 × 4 × 4 × 4 = 5 × 4 5 = 5120. 4. The total number of words that can be formed from 8 distinct letters is 8!. However, the A’s are not distinct, thus we need to reduce by a factor of 3!. The B’s are not distinct and we need to reduce by a factor of 2!. Thus, the total number is 8! / (3!2!) = 3360. 5. There are 4! = 24 ways in which the four volumes can be put back. Only one of these
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Unformatted text preview: is correct. P (correct order) = 1 4! = 1 24 . 6. This problem is similar to problem 4. First assume all the 7 heads and 3 tails are distinct. Then there are 10! ways. But the 7 heads are identical and the 3 tails are identical. Thus, the total number of ways is 10! / (7!3!) = 120. 7. Combinations of 1 out of 5 = ‡ 5 1 · = 5; combinations of 2 out of 5 = ‡ 5 2 · = 5! / 3!2! = 10; sum of all combinations = ∑ 5 j =0 ‡ 5 j · = 2 5 = 32. 8. Combinations of 5 out of 20 = ‡ 20 5 · . The number of ways of selecting 2 defectives out of 3 defectives and 3 non-defectives out of 17 non-defectives = ‡ 3 3 ·‡ 17 3 · . Thus, P2 defectives in 5 samples = ‡ 3 2 ·‡ 17 3 · / ‡ 20 5 · ....
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This note was uploaded on 06/09/2010 for the course EE 131A taught by Professor Lorenzelli during the Spring '08 term at UCLA.

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